Tension in cable = T
Force up at end from cable = T sin 37
compression Force toward wall = T cos 37
max friction force = .485 T cos 37 up
vertical forces on rod sum to 0
2 Fg = .485 T cos 37 + T sin 37
2 Fg = .989 T
T = 2.02 Fg
take moments about intersection of rod and wall
clockwise
(1.6 + x) Fg
counterclockwise
3.2 T sin 37
so
3.2 (2.02)Fg (.602) = (1.6+x) Fg
x + 1.6 = 3.89
x = 2.29
One end of a uniform 3.20-m-long rod of weight Fg is supported by a cable at an angle of θ = 37° with the rod. The other end rests against the wall, where it is held by friction as shown in the figure below. The coefficient of static friction between the wall and the rod is μs = 0.485. Determine the minimum distance x from point A at which an additional object, also with the same weight Fg, can be hung without causing the rod to slip at point A.
3 answers
where do you take 1.6
1.6 is half of 3.2 so half the length of rod
1 answer